Math Problem Statement

A tank that resembles a hemisphere flat face up is full of water, with a small outlet on the top right. The weight of water is 62.5 lb/ft^3, find the work required to pump the water out of the outlet.

Solution

To find the work required to pump the water out of a hemispherical tank with a small outlet at the top, we need to use the concept of work in the context of pumping liquids. The work WW is calculated by integrating the force exerted over a distance. Here's how we can solve this step-by-step:

1. Define the Problem:

  • Shape of the Tank: Hemisphere with the flat face up.
  • Weight of Water: 62.562.5 lb/ft³.
  • Radius of the Hemisphere: Let’s denote it as RR.

2. Determine the Differential Element:

Consider a thin horizontal slice of water at a height yy from the bottom of the tank, where the bottom of the hemisphere is at y=0y = 0 and the top of the hemisphere is at y=Ry = R.

  • Thickness of the slice: dydy.

  • Volume of the slice: This can be represented as the area of the circular cross-section at height yy times the thickness dydy.

    The area of the cross-section is given by the formula for the area of a circle with radius R2y2\sqrt{R^2 - y^2} (since the radius at height yy decreases as we move up).

    Area=π(R2y2)\text{Area} = \pi \left( R^2 - y^2 \right)

    So, the volume dVdV of the slice is:

    dV=π(R2y2)dydV = \pi (R^2 - y^2) \, dy

3. Calculate the Weight of the Slice:

The weight dWdW of this thin slice is given by the density of water times the volume:

dW=62.5π(R2y2)dydW = 62.5 \cdot \pi (R^2 - y^2) \, dy

4. Calculate the Work to Pump this Slice to the Top:

To find the work required to pump this slice out of the outlet at the top, we need to move it a distance of RyR - y. The work dWdW to move the slice is given by:

dWork=dW(Ry)=62.5π(R2y2)(Ry)dyd\text{Work} = dW \cdot (R - y) = 62.5 \cdot \pi (R^2 - y^2) (R - y) \, dy

5. Integrate Over the Entire Volume:

To find the total work, we integrate from y=0y = 0 to y=Ry = R:

W=0R62.5π(R2y2)(Ry)dyW = \int_0^R 62.5 \cdot \pi (R^2 - y^2) (R - y) \, dy

6. Evaluate the Integral:

Simplifying the integral:

W=62.5π0R(R3Ry2R2y+y3)dyW = 62.5 \pi \int_0^R (R^3 - R y^2 - R^2 y + y^3) \, dy

Let's break it down:

0R(R3Ry2R2y+y3)dy=[R3yR3y3R22y2+y44]0R\int_0^R (R^3 - R y^2 - R^2 y + y^3) \, dy = \left[ R^3 y - \frac{R}{3} y^3 - \frac{R^2}{2} y^2 + \frac{y^4}{4} \right]_0^R

Substitute y=Ry = R and y=0y = 0:

R3(R)R3(R)3R22(R)2+(R)44R^3(R) - \frac{R}{3} (R)^3 - \frac{R^2}{2} (R)^2 + \frac{(R)^4}{4} =R4R43R42+R44= R^4 - \frac{R^4}{3} - \frac{R^4}{2} + \frac{R^4}{4} Combine like terms:

R4(11312+14)=R4(1212412612+312)R^4 \left(1 - \frac{1}{3} - \frac{1}{2} + \frac{1}{4}\right) = R^4 \left(\frac{12}{12} - \frac{4}{12} - \frac{6}{12} + \frac{3}{12}\right) =R4(512)= R^4 \left(\frac{5}{12}\right)

Finally, multiply by the constant:

W=62.5πR4512=62.5πR42.4W = 62.5 \pi R^4 \cdot \frac{5}{12} = \frac{62.5 \pi R^4}{2.4}

So, the total work required is:

W=312.5πR412 lb-ftW = \frac{312.5 \pi R^4}{12} \text{ lb-ft}

This gives the amount of work needed to pump all the water out of the hemisphere.

Questions to Consider:

  1. How would the work change if the outlet was at the center of the flat face instead of the top right?
  2. What happens to the work required if the tank was not full?
  3. How would this problem differ if the tank was a full sphere?
  4. What if the density of the fluid was different?
  5. How does the shape of the tank influence the required work?

Tip:

Always break down the integral into simpler parts to avoid mistakes in complex expressions!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Work-Energy Principle
Volume of Revolution

Formulas

Work = Force * Distance
Volume of a thin slice of water: dV = π(R^2 - y^2) dy
Force = Weight Density * Volume
Work for a slice: dW = 62.5 * π(R^2 - y^2) (R - y) dy
Total Work: W = ∫_0^R 62.5 * π (R^2 - y^2) (R - y) dy

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

University Level (Calculus)